Fe2+ + CuSO4 = FeSO4 + Cu2+ is what type of reaction
The molar mass of Hg is 200.59 g/mol and of O is 16.00 g/mol. HgO consists of 1 Hg and 1 O atom, so the molar mass of HgO is 200.59 + 16.00 = 216.59 g/mol. To find the mass of mercury produced, we calculate the molar mass ratio of Hg to HgO: (200.59 g/mol Hg) / (216.59 g/mol HgO) = 0.926. When 16 grams of HgO are heated, the mass of mercury produced would be 16 g * 0.926 ≈ 14.82 grams.
To find the molar ratios of Hg and O in HgO, we need to first find the molar mass of HgO: Hg = 200.59 g/mol, O = 16.00 g/mol. HgO = 200.59 + 16.00 = 216.59 g/mol. Next, divide the mass of HgO given (44.5g) by the molar mass of HgO to find the number of moles. Then, use the mole ratio to determine the grams of mercury and oxygen formed.
The reactants in the equation are mercury II oxide (HgO). This compound will decompose into mercury (Hg) and oxygen (O2) during the reaction.
i can't really read that reaction but! if there is an element by itself in either the reactant or the product side of the reaction QUICKLY think REDOX because the oxidation states for ANY element is 0 by themselves and when they form compounds they either oxidize or reduce to their oxidation state hope i helped
The chemical equation for mercury(II) oxide is HgO. It is composed of one mercury (Hg) atom and one oxygen (O) atom.
The word to describe HgO in Lavoisier's reaction demonstrating the conservation of mass is "reactant." In this reaction, mercury(II) oxide (HgO) decomposes upon heating into mercury (Hg) and oxygen gas (O2). Lavoisier's experiment illustrated that the total mass of the reactants (HgO) equals the total mass of the products (Hg and O2), thereby confirming the principle of conservation of mass.
55.8 grams
The molar mass of Hg is 200.59 g/mol and of O is 16.00 g/mol. HgO consists of 1 Hg and 1 O atom, so the molar mass of HgO is 200.59 + 16.00 = 216.59 g/mol. To find the mass of mercury produced, we calculate the molar mass ratio of Hg to HgO: (200.59 g/mol Hg) / (216.59 g/mol HgO) = 0.926. When 16 grams of HgO are heated, the mass of mercury produced would be 16 g * 0.926 ≈ 14.82 grams.
combustion
To determine how many grams of mercury (Hg) are needed to produce 3.75 moles of oxygen (O₂), we first consider the balanced chemical reaction for the formation of oxygen from mercury(II) oxide (HgO): 2 HgO → 2 Hg + O₂. From this, we see that 2 moles of HgO produce 1 mole of O₂, which means 3.75 moles of O₂ require 7.5 moles of Hg. The molar mass of Hg is approximately 200.59 g/mol, so 7.5 moles of Hg would weigh about 1,504.43 grams (7.5 moles × 200.59 g/mol).
To find the molar ratios of Hg and O in HgO, we need to first find the molar mass of HgO: Hg = 200.59 g/mol, O = 16.00 g/mol. HgO = 200.59 + 16.00 = 216.59 g/mol. Next, divide the mass of HgO given (44.5g) by the molar mass of HgO to find the number of moles. Then, use the mole ratio to determine the grams of mercury and oxygen formed.
The reactants in the equation are mercury II oxide (HgO). This compound will decompose into mercury (Hg) and oxygen (O2) during the reaction.
i can't really read that reaction but! if there is an element by itself in either the reactant or the product side of the reaction QUICKLY think REDOX because the oxidation states for ANY element is 0 by themselves and when they form compounds they either oxidize or reduce to their oxidation state hope i helped
The chemical formula for mercury(II) oxide is HgO. It is composed of one mercury (Hg) atom and one oxygen (O) atom.
The chemical equation for mercury(II) oxide is HgO. It is composed of one mercury (Hg) atom and one oxygen (O) atom.
When mercury(II) oxide (HgO) is heated, it decomposes into mercury and oxygen. The atoms present in the compound at the start of the reaction are mercury (Hg) and oxygen (O), and these same atoms will be present at the end of the reaction, just in different molecular forms.
The formula for mercurous peroxide is Hg2O2.